We were given a homework question in vectors and I'm not sure how to solve it.
For question A, I have tried to make the vectors equal each other since they're collinear which makes them appear on the same line. So I got 2/3b = 1/2c. Now, I'm not sure how to continue from here. In the textbook, the answer says m = 4 and n = -3, however, I'm not exactly sure how to get to here. And also it says the there are infinite values for m and n that would make the vectors null. I don't completely understand the logic behind this.
For question B, I'm not sure how to even start. I just need a lead so I could understand the principle behind this.
Any sort of help would be greatly appreciated, Thanks a lot!
$$\frac{2}{3}\vec{b} = \frac{1}{2}\vec{c} \Longrightarrow \vec{b}=\frac{3}{4}\vec{c}$$
$$m\vec{c} + n\vec{b} = \vec{0}$$
$$m\vec{c} + \frac{3}{4}n\vec{c} = \vec{0}$$
$$\vec{c}(m+\frac{3}{4}n) = \vec{0}$$
If $\vec{c}$ is a non-zero vector, then $m + \frac{3}{4}n = 0$.
Solving for $m$ and $n$:
$$ \begin{cases} m+\frac{3}{4}n=0\\ \end{cases} \sim \begin{cases} m=-\frac{3}{4}n\\ n=n \end{cases}, m, n \in \mathbb{R} $$
Therefore there are infinitely many possible values for $m$ and $n$ in the set of real numbers, simply because we are given only 1 equation in a system involving 2 variables. If you try $n=-3$, then $m=4$. This answer is only one possible pair of values that satisfy the equation.
So with this in mind, the same could be applied for the second part of the question. Rewrite each vector in terms of $\vec{a}$ to substitute in the equation. Then you'll find that each coefficient can be written in terms of another coefficient since only 1 equation is given in the system of 3 variables. You'll find that $d$ can be $2$, causing $e$ to equal $0$, and $f$ to equal $-1$. But just like before, this tuple of values is an element of the entire solution set.